Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $p \neq 0$. $n = \dfrac{p - 4}{p - 9} \times \dfrac{p^2 - 16p + 63}{p - 4} $
Solution: First factor the quadratic. $n = \dfrac{p - 4}{p - 9} \times \dfrac{(p - 9)(p - 7)}{p - 4} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (p - 4) \times (p - 9)(p - 7) } { (p - 9) \times (p - 4) } $ $n = \dfrac{ (p - 4)(p - 9)(p - 7)}{ (p - 9)(p - 4)} $ Notice that $(p - 4)$ and $(p - 9)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ \cancel{(p - 4)}(p - 9)(p - 7)}{ \cancel{(p - 9)}(p - 4)} $ We are dividing by $p - 9$ , so $p - 9 \neq 0$ Therefore, $p \neq 9$ $n = \dfrac{ \cancel{(p - 4)}\cancel{(p - 9)}(p - 7)}{ \cancel{(p - 9)}\cancel{(p - 4)}} $ We are dividing by $p - 4$ , so $p - 4 \neq 0$ Therefore, $p \neq 4$ $n = p - 7 ; \space p \neq 9 ; \space p \neq 4 $